Question: $ F = \left[\begin{array}{rrr}0 & 1 & 3 \\ 1 & 0 & -2\end{array}\right]$ $ A = \left[\begin{array}{rr}5 & 4 \\ 2 & 5 \\ 4 & 5\end{array}\right]$ What is $ F A$ ?
Solution: Because $ F$ has dimensions $(2\times3)$ and $ A$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ F A = \left[\begin{array}{rrr}{0} & {1} & {3} \\ {1} & {0} & {-2}\end{array}\right] \left[\begin{array}{rr}{5} & \color{#DF0030}{4} \\ {2} & \color{#DF0030}{5} \\ {4} & \color{#DF0030}{5}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rr}{0}\cdot{5}+{1}\cdot{2}+{3}\cdot{4} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{0}\cdot{5}+{1}\cdot{2}+{3}\cdot{4} & ? \\ {1}\cdot{5}+{0}\cdot{2}+{-2}\cdot{4} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{0}\cdot{5}+{1}\cdot{2}+{3}\cdot{4} & {0}\cdot\color{#DF0030}{4}+{1}\cdot\color{#DF0030}{5}+{3}\cdot\color{#DF0030}{5} \\ {1}\cdot{5}+{0}\cdot{2}+{-2}\cdot{4} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{0}\cdot{5}+{1}\cdot{2}+{3}\cdot{4} & {0}\cdot\color{#DF0030}{4}+{1}\cdot\color{#DF0030}{5}+{3}\cdot\color{#DF0030}{5} \\ {1}\cdot{5}+{0}\cdot{2}+{-2}\cdot{4} & {1}\cdot\color{#DF0030}{4}+{0}\cdot\color{#DF0030}{5}+{-2}\cdot\color{#DF0030}{5}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}14 & 20 \\ -3 & -6\end{array}\right] $